Can the closure of isolated point set be uncountable
1
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Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.
Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?
real-analysis general-topology
asked Jan 6 at 9:12
PaulPaul
194111
$endgroup$
add a comment |
1
$begingroup$
Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.
Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?
real-analysis general-topology
asked Jan 6 at 9:12
PaulPaul
194111
$endgroup$
$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
1
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44
add a comment |
1
1
1
$begingroup$
Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.
Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?
real-analysis general-topology
asked Jan 6 at 9:12
PaulPaul
194111
$endgroup$
Let $Ssubset [0,1]$ be a countable set, $bar{S}$ is the closure of $S$. Can $bar{S}$ be uncountable? I think it can not be, but I can't find a proof.
Edit: I just realized that my previous question is trivial. But what I really mean is that suppose $S$ is countable and with isolated points can $bar{S}$ be uncountable?
real-analysis general-topology
real-analysis general-topology
asked Jan 6 at 9:12
PaulPaul
194111
asked Jan 6 at 9:12
PaulPaul
194111
edited Jan 6 at 9:35
Paul
asked Jan 6 at 9:12
PaulPaul
194111
asked Jan 6 at 9:12
PaulPaul
194111
asked Jan 6 at 9:12
PaulPaul
194111
194111
$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
1
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44
add a comment |
$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
1
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44
$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
1
1
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44
add a comment |
1 Answer
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Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.
answered Jan 6 at 9:14
twnlytwnly
1,2211214
$endgroup$
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
add a comment |
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1 Answer
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1 Answer
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3
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Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.
answered Jan 6 at 9:14
twnlytwnly
1,2211214
$endgroup$
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
add a comment |
3
$begingroup$
Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.
answered Jan 6 at 9:14
twnlytwnly
1,2211214
$endgroup$
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
add a comment |
3
3
3
$begingroup$
Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.
answered Jan 6 at 9:14
twnlytwnly
1,2211214
$endgroup$
Closure of the rationals $mathbb{Q} cap [0,1]$ is $mathbb{R} cap [0,1]$, which is uncountable.
answered Jan 6 at 9:14
twnlytwnly
1,2211214
answered Jan 6 at 9:14
twnlytwnly
1,2211214
answered Jan 6 at 9:14
twnlytwnly
1,2211214
answered Jan 6 at 9:14
twnlytwnly
1,2211214
1,2211214
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
add a comment |
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
$begingroup$
Yeah... When I submited the question, I realized it is trivial...
$endgroup$
– Paul
Jan 6 at 9:19
add a comment |
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$begingroup$
What is the closure of the set $$S={(frac mn,frac1n):minmathbb Z,ninmathbb N}$$ in $mathbb R^2$?
$endgroup$
– bof
Jan 6 at 9:52
$begingroup$
Do you mean that it has every point isolated or just some? If the latter, consider $[0,1]cap Bbb Q)cup{17}$.
$endgroup$
– MJD
Jan 6 at 10:56
1
$begingroup$
In $mathbb R$ let $$S=left{frac12,frac16,frac56,frac1{18},frac5{18},frac{13}{18},frac{17}{18},frac1{54},dotsright},$$ the set of midpoints of the intervals removed in constructing the Cantor set; $S$ is a discrete set (every point is isolated), and its set of limit points is the Cantor set, which is uncountable.
$endgroup$
– bof
Jan 6 at 11:35
$begingroup$
@bof Nice example!
$endgroup$
– Paul
Jan 6 at 11:44